Is the derivative of a Lipschitz function better than L^\infty - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T17:28:07Z http://mathoverflow.net/feeds/question/57386 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/57386/is-the-derivative-of-a-lipschitz-function-better-than-l-infty Is the derivative of a Lipschitz function better than L^\infty Phil Isett 2011-03-04T19:02:02Z 2011-03-04T23:19:09Z <p>How smooth is the first derivative (in the distribution sense) of a Lipschitz function? Taking difference quotients and testing against an $L^1$ function, we see that $Df$ is in $L^\infty$. In ${\mathbb R}^1$ the converse is true, thanks to the persistence of the formula</p> <p>$f(x+h) - f(x) = \int_0^1 f'(x+th) dt~ h$</p> <p>(Proof: convolve with a mollifier)</p> <p>However, if $f : {\mathbb R}^n \to {\mathbb R}$ is Lipschitz, then by the same argument, its derivative has a restriction to any line which is in $L^\infty$ of that line (more precisely, the tangential component of the derivative restricts). Ordinarily, one cannot restrict a distribution sensibly to lower dimensional subsets (straight lines requiring even more regularity than curves), or at least if you can because its primitive restricts, I don't know of any reason to expect the restriction to have any semblance of regularity.</p> <p>For $n > 1$, is there a nice Banach space in which the derivative of a Lipschitz function belongs whose elements are smoother than just $L^\infty$?</p> http://mathoverflow.net/questions/57386/is-the-derivative-of-a-lipschitz-function-better-than-l-infty/57408#57408 Answer by Tom LaGatta for Is the derivative of a Lipschitz function better than L^\infty Tom LaGatta 2011-03-04T22:09:23Z 2011-03-04T22:09:23Z <p>Every Lipschitz function is absolutely continuous. Consequently, its derivative exists and is uniformly bounded almost everywhere. The Lipschitz constant is just the $L^\infty$ norm of the derivative.</p> <p>If you want a Banach space of smoother functions, then just define it. For example, let $X$ be the space of Lipschitz functions on $\mathbb R^n$ with integrable derivatives: <code>$$X = \{ f :~ \nabla f \in L^1 \cap L^\infty \}.$$</code></p> http://mathoverflow.net/questions/57386/is-the-derivative-of-a-lipschitz-function-better-than-l-infty/57418#57418 Answer by Spencer for Is the derivative of a Lipschitz function better than L^\infty Spencer 2011-03-04T23:19:09Z 2011-03-04T23:19:09Z <p>Lipschitz functions are exactly $W^{1,\infty}$ (See '<a href="http://en.wikipedia.org/wiki/Sobolev_space" rel="nofollow">Sobolev space</a>' on wikipedia - under other examples and perhaps the bit about absolute continuity on lines). This means the short answer to your question is no.</p>